Problem: Let $f(x) = -8x^{2}+10x-3$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Answer: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-8x^{2}+10x-3 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -8, b = 10, c = -3$ $ x = \dfrac{-10 \pm \sqrt{10^{2} - 4 \cdot -8 \cdot -3}}{2 \cdot -8}$ $ x = \dfrac{-10 \pm \sqrt{4}}{-16}$ $ x = \dfrac{-10 \pm 2}{-16}$ $x =\frac{1}{2},\frac{3}{4}$